Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is 
above the ground
Using the equation of motion we can write
Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .
The correct answer is: <span>Unscrew one light, if the others remain on it is a parallel circuit.</span>
Answer:
Explanation:
To solve this, we start by using one of the equations of motion. The very first one, in fact
1
V = U + at.
V = 0 + 0.8 * 3.4 = 2.72 m/s.
2.
V = 0 + 0.8 * 4.3 = 3.44 m/s.
3.
d = ½ * 0.8 * 4.3² + 3.44 * 12.9
d = 7.396 + 44.376
d = 51.77 m.
4.
d = 62 - 51.77 = 10.23 m. = Distance
traveled during deceleration.
a = (V² - Vo²) / 2d.
a = (0² - 3.44²) / 20.46
a = -11.8336 / 20.46 = -0.58 m/s²
5.
t = (V - Vo)/a =(0 - 3.44) / -0.58
t = -3.44/-.58 = 5.93 s
= Stop time.
T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total
time the hare was moving.
6.
d = Vo * t + ½ * a * t² = 62 m.
0 + 0.5 * (23.13)² * a = 61
267.5a = 61
a = 61/267.5
a = 0.23 m/s²
Answer:
only the weight of the ball will act on the ball
Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown
